A Treatise of Practical Mathematics, Part 2W. & R. Chambers, 1842 |
From inside the book
Results 1-5 of 47
Page 9
... contain respectively the right and left offsets . The station from which the measurements are begun is called the first station ; that next arrived at , the second ; and so on . The field - book of the measurements of a field similar to ...
... contain respectively the right and left offsets . The station from which the measurements are begun is called the first station ; that next arrived at , the second ; and so on . The field - book of the measurements of a field similar to ...
Page 15
... contained by these distances . The field is thus divided into triangles , in each of which two sides and the contained angle are known ; and their areas may therefore be found by article 259 in Mensuration of Surfaces , and their sum ...
... contained by these distances . The field is thus divided into triangles , in each of which two sides and the contained angle are known ; and their areas may therefore be found by article 259 in Mensuration of Surfaces , and their sum ...
Page 16
... contained angle in each triangle ; and hence their areas can be found , the sum of which is that of the given field . - EXERCISE . Find the area of the subjoined field ABCDE , from these measurements : - Side AB = 388 Angle B = 110 ° 30 ...
... contained angle in each triangle ; and hence their areas can be found , the sum of which is that of the given field . - EXERCISE . Find the area of the subjoined field ABCDE , from these measurements : - Side AB = 388 Angle B = 110 ° 30 ...
Page 23
... contain 1 acre 2 roods 16 poles , by a line parallel to one of its ends ; what is the length of this part ? Ans . 500 links . 37. PROBLEM X. - To cut off any portion from a triangle by a line drawn from its vertex . Let ABC be the ...
... contain 1 acre 2 roods 16 poles , by a line parallel to one of its ends ; what is the length of this part ? Ans . 500 links . 37. PROBLEM X. - To cut off any portion from a triangle by a line drawn from its vertex . Let ABC be the ...
Page 24
... contain 2 acres I rood 10 poles . A5 ac . 2 ro . 15 pls . - 559375 links a = 2 ac . 1 ro . 10 pls . = 231250 Hence , α S√1525 √ 231250 559375 ... = 980-5 lks . = AD . EXERCISE . The area of a triangle is 12.96 acres 24 PRACTICAL ...
... contain 2 acres I rood 10 poles . A5 ac . 2 ro . 15 pls . - 559375 links a = 2 ac . 1 ro . 10 pls . = 231250 Hence , α S√1525 √ 231250 559375 ... = 980-5 lks . = AD . EXERCISE . The area of a triangle is 12.96 acres 24 PRACTICAL ...
Contents
207 | |
209 | |
218 | |
225 | |
234 | |
241 | |
247 | |
259 | |
95 | |
103 | |
111 | |
112 | |
119 | |
127 | |
129 | |
137 | |
144 | |
151 | |
165 | |
172 | |
178 | |
187 | |
192 | |
200 | |
206 | |
267 | |
273 | |
283 | |
291 | |
299 | |
309 | |
316 | |
329 | |
336 | |
338 | |
344 | |
352 | |
363 | |
370 | |
376 | |
382 | |
389 | |
Common terms and phrases
actual range bastion body breadth bung diameter called cask centre constructed content in imperial cosec cosine course declination deflexion depth dial difference of latitude difference of longitude distance ditch divide divisor draw earth elevation equal EXAMPLE EXERCISES fathoms Find the content flank frustum gauge point glacis hence horizontal hypotenuse imperial bushels imperial gallons impetus inches inclination length logarithms longitude mean diameter measure meridian middle moon's multiply number of balls oblique observed opposite parallax parallel parapet perpendicular pile place of arms plane plane sailing pole potential range preceding primitive PROBLEM projectile projection proportional quadrant quotient radius ravelin redoubt refraction right angle right ascension sailing semi-diameter Severndroog Castle side sidereal sin² sine Sliding Rule small circle specific gravity spherical angle spherical excess spherical triangle spherical trigonometry square sun's surface true altitude ullage velocity vessel weight
Popular passages
Page 196 - Fig. 9. Case 1. Let AB, AC be each less than a quadrant. Let AE, AG be quadrants ; G will be the pole of AB, and E the pole of AC, and EC a quadrant; but, by prop. 12. CE is greater than CB, since CB is farther off from CGD than CE. In the same manner, it is shown...
Page 95 - To the square of the bung diameter add the square of the head diameter ; multiply the sum by the length, and the product again by .0014 for ale gallons, or by .0017 for wine gallons.
Page 96 - RULE. — To the square of the bung diameter add the square of the head diameter ; multiply the sum by the length, and the product by .0014 for ale gallons, or by .0017 for wine gallons.
Page 42 - A magnitude which has length, breadth, and thickness. Solution. The process by which the answer to a question is obtained. Specific gravity of a substance. The ratio of the weight of a given volume of it to that of an equal volume of water.
Page 192 - A sphere is a solid, bounded by one continued convex surface, every point of which is equally distant from a point within, called the centre. The sphere may be conceived to be formed by the revolution of a semicircle about its diameter, which remains fixed.
Page 227 - ZODIAC.— The Zodiac is an imaginary belt, or broad circle, extending quite around the heavens. The ecliptic divides the zodiac into two equal parts, the zodiac extending 8 degrees on each side of the ecliptic, and therefore is 16 degrees wide.
Page 196 - BC will be greater than a quadrant : for let AE be a quadrant, then E is the pole of AC, and EC will be a quadrant. But CB is greater than CE by Prop. 12.
Page 195 - Oj the same affection with the angles opposite to them, that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles, and conversely.
Page 195 - IN a right angled spherical triangle, the sides are of the same affection with the opposite angles ; that is, if the sides be greater or less than quadrants, the opposite angles will be greater or less than right angles. Let ABC be a spherical triangle right angled at A, any side AB, will be of the same affection with the opposite angle ACB. Case 1.
Page 195 - ... will be greater than a quadrant. Let ABC be a right angled spherical triangle ; according as the two sides AB, AC are of the same or of different affection, the hypotenuse BC will be less, or greater than a quadrant. The...